Let's go through the problem in number 9 step by step. Here's what it asks:

Given table:

Quantity Sold and Produced	Costs	Revenue	Profit
600	$9,350	$25,000	$15,650
1200	$18,350	$45,600	$27,250
1800	$27,350	$57,650	$30,300
3000	$45,360	$60,000	$14,640

Now, let’s solve the subquestions one by one:

9a) Fill in the table and identify at which level profit is maximized.

We are asked to identify the quantity of items where profit is maximized based on the table.

• Profit = Revenue - Costs
• The profits are already calculated and filled in for each quantity level:
  • For 600 units: Profit = $25,000 - $9,350 = $15,650
  • For 1200 units: Profit = $45,600 - $18,350 = $27,250
  • For 1800 units: Profit = $57,650 - $27,350 = $30,300
  • For 3000 units: Profit = $60,000 - $45,360 = $14,640

Profit is maximized at 1800 units sold, where the profit is $30,300, as correctly indicated.

9b) What is the average rate of change in Profit between 1200 and 1800 units sold?

The formula for the average rate of change is: $\text{Average rate of change} = \frac{\Delta \text{Profit}}{\Delta \text{Quantity}}$ Here: $\Delta \text{Profit} = 30,300 - 27,250 = 3,050$ $\Delta \text{Quantity} = 1800 - 1200 = 600$ So, the average rate of change in profit is: $\frac{3050}{600} = 5.08$ This means the profit increases by $5.08 for each additional unit sold between 1200 and 1800 units.

9c) Find a quadratic regression equation to model profit and the associated marginal profit function.

You are asked to model the profit as a quadratic function $P(x)$ based on the given data points. The profit values at $x = 600, 1200, 1800, 3000$ can be used to estimate this quadratic function of the form: $P(x) = ax^2 + bx + c$ Where:

• $P(600) = 15650$
• $P(1200) = 27250$
• $P(1800) = 30300$
• $P(3000) = 14640$

From your work and based on the hint about Desmos, you already found the regression equation: $P(x) = 3050 - 1200x^2 + 1800x$ This quadratic function models how the profit changes with the number of units sold.

The marginal profit function is the derivative of $P(x)$. If $P(x) = ax^2 + bx + c$, then: $P'(x) = 2ax + b$ Substituting the values for $a$ and $b$ from the regression equation, you would get the marginal profit function.

9d) Profit is maximized when $x = 4$.

It seems you calculated that the maximum profit occurs at $x = 4$. Without further clarification, this likely refers to a specific point on a scaled graph that was obtained through further graphing or algebraic work.

Would you like further clarification on this part, or should I help derive the derivative for marginal profit directly?

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Let me know if you would like any additional details or clarifications. Here are 5 related questions to consider next:

1. How is the quadratic regression formula derived step-by-step for a dataset like this?
2. What does the marginal profit tell us about how profit changes with each additional unit sold?
3. Can you explain how second-order derivatives can help in determining whether profit is maximized?
4. How can profit models like this be used in real business decision-making processes?
5. What other forms of regression (e.g., linear) could we use to model different types of business data?

Tip: Understanding the relationship between cost, revenue, and profit can help businesses maximize profitability by finding the optimal balance between production and pricing strategies.